Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation MnO 2 (s)+4 HCl(aq) longrightarrow MnCl 2 (aq)+2 H 2 O(l)+Cl 2 (g) How much Mn*O_{2}(s) should be added to excess HCl(aq) to obtain 355 mL C*l_{2}(g) at 25 deg * C and 785 Torr? 9

Answer:

Calculating the Amount of Manganese Dioxide Needed to Produce 355mL of Chlorine Gas

The reaction of manganese dioxide and hydrochloric acid, known as the Chlorine Gas Law, can be used to calculate the amount of manganese dioxide needed to produce a certain volume of chlorine gas. In order to produce 355 mL of chlorine gas at 25 degrees Celsius and 785 Torr, the following equation must be used:

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g).

Calculation of the Amount of MnO2 Needed

In order to calculate the amount of manganese dioxide needed to produce 355 mL of chlorine gas, the Ideal Gas Law must be used. The Ideal Gas Law states that the volume of a gas is equal to the number of moles of the gas multiplied by the ideal gas constant, R, divided by the pressure of the gas. The equation is as follows:

V = nR/P

In this case, V is equal to 355 mL, n is equal to the moles of chlorine gas produced, R is equal to 0.0821 L atm/molK and P is equal to 785 Torr. Therefore, the equation becomes:

355 mL = n(0.0821 L atm/molK)/785 Torr

By solving for n, the moles of chlorine gas produced can be determined.

n = 355 mL / (0.0821 L atm/molK / 785 Torr) = 0.0045 mol

Calculation of the Amount of MnO2 Needed to Produce 0.0045 mol of Cl2

Now that the moles of chlorine gas produced are known, the amount of manganese dioxide needed to produce 0.0045 mol of chlorine gas can be determined. The equation for the reaction of manganese dioxide and hydrochloric acid is as follows:

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g).

To produce 0.0045 mol of chlorine gas, 4 times the amount of moles of hydrochloric acid, or 0.018 mol HCl, must be used. The equation becomes:

MnO₂(s) + 0.018 mol HCl(aq) → MnCl₂(aq) + 2H₂O(l) + 0.0045 mol Cl₂(g).

The equation can be solved for MnO₂(s) by dividing both sides of the equation by 0.018 mol HCl(aq).

MnO₂(s) = 0.0045 mol Cl₂(g) / 0.018 mol HCl(aq)

Therefore, the amount of manganese dioxide needed in order to produce 355 mL of chlorine gas at 25 degrees Celsius and 785 Torr is 0.0045 mol Cl₂(g) / 0.018 mol HCl(aq) = 0.25 mol MnO₂(s).

Related Questions

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